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Question Number 132257 by Salman_Abir last updated on 12/Feb/21

Answered by bemath last updated on 12/Feb/21

(i)E(0,0,8), C(3,3,0) vector EC = 3i^� +3j^� −8k^� ; ∣EC∣=(√(9+9+64)) =(√(82)) H(0,3,8) ,B(3,0,0) vector HB=3i^� −3j^� −8k^� ; ∣HB∣ = (√(82)) so ∣EC∣=∣HB∣ but EC ≠ HB (ii) vector HG = { ((EF ; AB)),((DC )) :} (iii) EF+BD+CF=AB+BD+CF=BF

$$\left(\mathrm{i}\right)\mathrm{E}\left(\mathrm{0},\mathrm{0},\mathrm{8}\right),\:\mathrm{C}\left(\mathrm{3},\mathrm{3},\mathrm{0}\right)\:\mathrm{vector}\: \\ $$$$\:\mathrm{EC}\:=\:\mathrm{3}\hat {\mathrm{i}}+\mathrm{3}\hat {\mathrm{j}}−\mathrm{8}\hat {\mathrm{k}}\:;\:\mid\mathrm{EC}\mid=\sqrt{\mathrm{9}+\mathrm{9}+\mathrm{64}}\:=\sqrt{\mathrm{82}} \\ $$$$\:\mathrm{H}\left(\mathrm{0},\mathrm{3},\mathrm{8}\right)\:,\mathrm{B}\left(\mathrm{3},\mathrm{0},\mathrm{0}\right)\:\mathrm{vector}\: \\ $$$$\:\mathrm{HB}=\mathrm{3}\hat {\mathrm{i}}−\mathrm{3}\hat {\mathrm{j}}−\mathrm{8}\hat {\mathrm{k}}\:;\:\mid\mathrm{HB}\mid\:=\:\sqrt{\mathrm{82}} \\ $$$$\:\mathrm{so}\:\mid\mathrm{EC}\mid=\mid\mathrm{HB}\mid\:\mathrm{but}\:\mathrm{EC}\:\neq\:\mathrm{HB}\: \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{vector}\:\mathrm{HG}\:=\:\begin{cases}{\mathrm{EF}\:;\:\mathrm{AB}}\\{\mathrm{DC}\:}\end{cases} \\ $$$$\left(\mathrm{iii}\right)\:\mathrm{EF}+\mathrm{BD}+\mathrm{CF}=\mathrm{AB}+\mathrm{BD}+\mathrm{CF}=\mathrm{BF} \\ $$

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